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nirvana33 [79]
3 years ago
15

Line M passes through point (3,1) and is perpendicular to the line of the equation y=3x+4. Which equation describes the graph of

M?
A) y=-3x+16
B) y=-1/3x+2
C)y=-1/3x+4
D)y=1/3x+6
Mathematics
1 answer:
zheka24 [161]3 years ago
8 0

Answer:

y = -1/3x + 2

Step-by-step explanation:

The gradient of the given line is 3 because (y = mx +c where m is the gradient)

Therefore, to find the gradient of the perpendicular line (at 90 degrees), you need to find the negative reciprocal.

The negative reciprocal of 3 is -1/3 because imagine if 3 = 3/1, to get the reciprocal, you flip it, and to get the negative, you just flip the sign.

Now we know that Line M is y = -1/3x + c, we need to find the y-intercept.

To do this, just input the point (3,1) into y = -1/3x + c, to get c. This is because we know (3,1) is on the line from the question.

So it would be 1 = (-1/3 x 3) +c

Which would be 1 = -1 +c

And so c = 2

Put everything together and you get y = -1/3x + 2

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Read 2 more answers
100 POINTSSS! ASAP ANSWEERR PLS
Margarita [4]

PART A

Given:

f(x) = 0.69(1.03)x

To find:

If the price of the product is increasing or decreasing and by what percentage

Steps:

we know the formula to find the price of Product A per year, so

f(1) = 0.69 * 1.03 * 1

Price = $0.7107

f(2) = 0.69 * 1.03 * 2

Price = $1.4214

Here the Price of Product after 2 years is greater than the price of Product after one year. So the price of the product A is increasing.

Now to find percentage increase,

Percentage increase = \frac{FV-SV}{SV}*100        (FV = final value, SV = starting value)

Percentage increase = \frac{1.4214 - 0.7107}{0.7107}*100

Percentage increase = \frac{0.7107}{0.7107}*100

Percentage increase = 100 %

Therefore, the percentage increase of Product A is 100%

PART B

Given:

Price of product B in 1st year = $10,100

Price of product B in 2nd year = $10,201

Price of product B in 3rd year = $10,303.01

Price of product B in 4th year = $10,406.04

To find:

Which product recorded a greater percentage change over the previous year

Steps:

We need to find the percentage change of Product B and Product A of each year. We know that the percentage change of product A is 100 % for each year, so we only need to calculate for product B

PC of product B from 1st to 2nd year = \frac{10,201-10,100}{10,100}*100

                                                             = \frac{101}{10,100}*100

                                                             = 0.01 * 100

                                                             = 1 %

PC of product B from 2nd to 3rd year = \frac{10,303.01-10,201}{10,201} *100

                                                              = 1%

PC of product B from 3rd to 4th year =\frac{10,406.04-10,303.01}{10,303.01}*100

                                                              ≈ 1%

So, percentage change of product B is 1% per year

Therefore, Product A has greater percentage change

Happy to help :)

If u need more help, feel free to ask

6 0
3 years ago
Help me with #4 pleasedeeee
Romashka [77]

Answer:

\dfrac{dy}{dx}=-\dfrac{2x-y-1}{2y-x-1}

Step-by-step explanation:

Differentiating, you get ...

... x·dy +y·dx +dx +dy = 2x·dx +2y·dy

Collecting terms gives ...

... dy(x +1 -2y) = dx(2x -y -1)

... dy/dx = -(2x -y -1)/(2y -x -1)

4 0
3 years ago
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